Simplifying Summation with Logarithms

[thereddevils]

Homework Statement

$$\sum^{n}_{r=1}(\lg \frac{2^r(r+1)}{r})=\sum^{n}_{r=1}[\lg 2^r+\lg (r+1)-\lg r]$$

Homework Equations

The Attempt at a Solution

i found the answer to be $$(2^n-1)\lg 2 +\lg (n+1)$$

Am i correct , or it can be further simplified ? Thanks .

 

[tiny-tim]

Hi thereddevils!

Your lg(n+1) is fine, but the two-y part isn't …

how did you manage to get a 2ⁿ outside the lg ? ​

 

[thereddevils]

Hi thereddevils!

Your lg(n+1) is fine, but the two-y part isn't …

how did you manage to get a 2ⁿ outside the lg ? ​

Thanks , tiny tim !

oh , \sum^{n}_{r=1}\lg 2^r=\lg 2+2\lg 2+3\lg 2 +...

=\lg 2 (1+2+3+...)

=\frac{n}{2}(n+1)\lg 2

i think i see my mistake , is this correct ?

 

[tiny-tim]

(have a sigma: ∑ and try using the X² tag just above the Reply box )

Yes, that's all correct now!